$\int^{3}_{1}\dfrac{x}{x^2+1}\,dx\, = $
Answer: Strategy Let's first find the indefinite integral $\int\dfrac{x}{x^2+1}\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{x}{x^2+1}\,dx\,$, we can use U-substitution. If we let $ {u=x^2+1}$, then ${du=2x \, dx}$ and ${ x\,dx=\dfrac{1}{2}\, du}$. So we have: $\begin{aligned}\int\dfrac{x}{x^2+1}\,dx\,&=\int\dfrac{1}{{x^2+1}}\,\cdot {x\,dx}\,\,\\\\\\\\ &=\int\dfrac{1}{ u}\,\cdot {\dfrac12\, du}\,\\\\\\\\ &=\dfrac12\int\dfrac{1}{u}\,du\\\\\\\\ &=\dfrac{1}{2}\ln|u|+C\\\\\\\\ &=\dfrac{1}{2}\ln|x^2+1|+C\\\\\\\\ &=\dfrac{1}{2}\ln(x^2+1)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{3}_{1}\dfrac{x}{x^2+1}\,dx\, &= \dfrac{1}{2}\ln(x^2+1)\Bigg|^3_1\\\\\\\\ &=\dfrac{1}{2}\left(\ln(10)-\ln(2)\right)\\\\\\\\ &=\dfrac12\ln(5)\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{3}_{1}\dfrac{x}{x^2+1}\,dx\, = \dfrac12\ln(5)$